# Mutually exclusive vs. independent

I understand that mutually exclusive events have no intersection, and this can be shown on a Venn diagram as separate sets. And I understand that independent variables are such that $$P(AB) = P(A)P(B)$$, so if $$P(AB) = 0$$ then $$P(A) = 0$$ or $$P(B) = 0$$. But can you show independent variables on a Venn diagram? Why is it that $$P(AB)$$ isn't always zero, if A and B are independent?

Consider flipping a single coin. Then $$P(heads) = 1/2$$ and $$P(tails) = 1/2$$. However, $$P(HT) = 0$$, since it is impossible to show heads and tails simultaneously. So I would conclude that H and T are not independent in this case. And would you say that H and T are mutually exclusive?

Now consider flipping two separate coins, again with $$P(H) = P(T) = 1/2$$. This time $$P(HT) = 1/4$$, as the coins are independent. They do not influence each other. But are the coins mutually exclusive? It seems they can't be, since $$P(HT)$$ is not zero.

I'm getting confused about mutually exclusive vs. independence. Please provide examples of mutually exclusive and independent, mutually exclusive and dependent, not mutually exclusive and independent, and not mutually exclusive and dependent.

• Venn diagrms do not take care of probbiblities. Independence involves probabilities. – Kavi Rama Murthy Apr 25 at 5:15

Mutually exclusive events are by definition dependent. Think about it: if I flip one coin and consider two events,

• $$H$$: The coin flipped shows heads.
• $$T$$: The coin flipped shows tails.

Each one of these have $$P(H) = P(T) = 1/2$$. Now consider the event $$HT$$ (that is, $$H\wedge T$$; $$H$$ AND $$T$$). Is it possible that one coin shows both heads and tails? No, hence $$P(HT) = 0$$. Since $$P(HT) \neq P(H)P(T)$$, $$H$$ and $$T$$ are dependent events. As a matter of fact they are so dependent that knowing one will immediately tell you the other (if I know that the coin shows heads, I immediately know that $$H$$ is true and that $$T$$ is false).

Now consider instead throwing two times the same coin and consider the events

• $$H_1$$: The first coin shows heads,
• $$H_2$$: The second coin shows heads,

Again we have that $$P(H_1) = P(H_2) = 1/2$$. However, now the event $$H_1H_2$$ happens one in four cases, so $$P(H_1H_2) = 1/4 = P(H_1)P(H_2)$$: the events are independent (and not mutually exclusive)!

Again consider tossing two times the same coin and consider the events

• $$T_1$$: The first coin shows tails.
• $$S$$: I see two heads in total.

We have that $$P(T_1) = 1/2$$ and $$P(S) = P(H_1H_2) = 1/4$$. In this case, $$S$$ and $$T_1$$ are mutually exclusive as if $$T_1$$ is true one coin shows tails and then I cannot see two heads and $$S$$ has to be false (so $$P(T_1 S) = 0$$). In addition, they are dependent, as $$P(T_1S) = 0 \neq P(T_1)P(S)$$.

At last, consider the events

• $$H_1$$: The first coin shows heads.
• $$S$$: I see two heads in total.

In this case the events $$H_1$$ and $$S$$ are not mutually exclusive; however, they are dependent as $$P(H_1 S) = P(S) = 1/4 \neq P(H_1)P(S)$$.